Finding the probability of the disease

Question

Only 0.01% of people have triskaidekaphobia. The Dreizehn Club has developed a test for the phobia. If you have Triskadekaphobia, the test is 99% likely to identify that you have the disease. Unfortunately, it is also 5% likely to claim you have the disease when you don’t. If you take the test twice and it twice claims you have the disease, what is the probability that you do have the disease?

I have tried solving the problem using Bayes' theorem however my calculated answer seems to be incorrect. The answer for the problem is 0.0377.

Here's what I tried so far: Let P(D) be the probability of having the disease and P(T) be the probability of the test being positive. P(T) = P(T|D)*P(D) + P(T|!D)*P(!D)

Pr(D|$T^2$)=$\frac{Pr(T^2|D)Pr(D)}{Pr(T^2)}$

Why does this not work?

Answer 1

Most probably you might have miscomputed $Pr(T^2)$

Since we are computing probability (a ratio), we can as well use the $\%$ figures directly,
if you aren't getting it, try

$\dfrac{0.01\times99^2}{0.01\times99^2 + 99.99\times5^2}$

Answer 2

$P(D)=0.0001$

$P(T_n\mid D) = .99, P(T_n\mid D')=0.05$

Find $P(D\mid T_1, T_2) = \dfrac{\overline{\underline{|\qquad\qquad|}}}{\overline{\underline{|\qquad\qquad|}}+\overline{\underline{|\qquad\qquad|}}} \approx 0.037{\small 7}$